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Stats - Z Test watch

    • Thread Starter

    In a school, a standardized reading test is used to test the performance of students againast the national norm for that age group. The number of students taking the test in this school is 55.

    The national norm score is 100, with a standard deviation for this year of 12. The school learns that the mean for their students is 96. Is the school's mean test score sufficiently lower than the national norm as to indicate a problem?
    Testing at 5% confidence level.

    H_0 : \bar{x} = \mu

    H_1 : \bar{x} < \mu

    \mathrm{SE} = \frac{12}{\sqrt{55}}

    z = \frac{\sqrt{55}}{3} = -2.47

    (*)\,\mathrm{P}(\mathrm{Z} \leq -2.47) = 1 - \mathrm{P}(\mathrm{Z} \leq 2.47) = 1 - 0.9932 = 0.0068 < 0.05

    We reject H_0 as there is evidence to suggest that the school's mean is not insignificantly lower than the national average.

    Is this right? Particularly (*) w.r.t. one- or two-tailed test.

    (It seems odd rejecting H_0 for H_1 given that they're not exhaustive - shouldn't H_0 : \bar{x} \leq \mu or something? This is certainly how the textbook does it, though - although this particular question is an adaptation of something found on Wikipedia.)
    • Study Helper

    Study Helper
    Seems right to me, although I would have said:

    Therefore we reject H_0.

    So yes, there is evidence to suggest that the school's mean is sufficiently lower than the national norm to indicate a problem.

    Rather than use a double negative, and this also ties it back to the original question.

    H_0 \text{ and }H_1 are merely alternative viewpoints, and there is no need for them to be exhaustive. The first is assumed to be true, unless statistically significant evidence shows otherwise.

    That's probably no help, but don't know what else I can say on it.
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Updated: January 28, 2009

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